|
Posted by peter on 08/28/07 15:17
> It returns null, which brings up a blank image.
then you can simply do something such as:-
<?PHP
$image = getRandomImage('LL/');
if (is_null($image))
{
$image = '';// Set to the image you wish
}
echo "<img src='$image' height=90 width=120 alt='LL'>";
?>
in the if statement place the path and name of the image you want to use as
the image that will be displayed if nothing is returned from the function.
Navigation:
[Reply to this message]
|